
A small mirror of mass $m$ is suspended by a light thread of length $l$. A short pulse of laser falls on the mirror with energy $E$. Then, which of the following statements is correct?

A) If the pulse falls normally on the mirror, it deflects by $\theta = \dfrac{{2E}}{{\left( {mc\sqrt {2gl} } \right)}}$.
B) If the pulse falls normally on the mirror, it deflects by $\theta = \dfrac{{2E}}{{\left( {mc\sqrt {2g} } \right)}}$.
C) Impulse in thread depends on angle at which the pulse falls on the mirror.
D) None of the above
Answer
126.6k+ views
Hint: In order to get the solution of the given question, we need to find the change in momentum of the laser beam. After that, we need to relate it with conservation of momentum and conservation of energy. After comparing, we will get an equation and solving the equation will give us the required solution of the given question.
Complete step by step solution:
The energy of the laser that falls on the mirror is given as $E$.
If we assume the laser beam to be perfectly reflected, then the change in momentum of laser beam is known as $\dfrac{{2E}}{c}.$
Now, if we apply conservation of momentum, we can write it as,
$mv = \dfrac{{2E}}{c}$
$ \Rightarrow v = \dfrac{{2E}}{{mc}}$
As, we know that from conservation of energy, Kinetic energy = Potential energy
Therefore, we can write, $\dfrac{1}{2}m{v^2} = mg(1 - \cos \theta )$………… (i)
Also, from trigonometry, we know that,$1 - \cos \theta = 2{\sin ^2}\dfrac{\theta }{2}$
Now, since the angle is very small, we can write the above relation as, $1 - \cos \theta = 2{\left( {\dfrac{\theta }{2}} \right)^2}$=$\dfrac{{{\theta ^2}}}{2}$
Now, using the above relation in equation (i), we can write it as,
$\dfrac{1}{2}m{\left( {\dfrac{{2E}}{{mc}}} \right)^2} = mgl\dfrac{{{\theta ^2}}}{2}$
$ \Rightarrow {\theta ^2} = \dfrac{{2 \times 2{E^2}}}{{c{m^2}gl}}$
$\therefore \theta = \dfrac{{2E}}{{mc\sqrt {gl} }}$
As we can notice, the required value of the angle is not given in the options of the given question.
Hence, option (D), i.e. none of the above is the correct choice of the given question.
Note: We define a laser beam as a ray of light with high intensity that is generated through the emission of photons (also known as light particles). Photons do not have mass but they are capable of redirecting particles with which they collide and transfer their momentum to them.
Complete step by step solution:
The energy of the laser that falls on the mirror is given as $E$.
If we assume the laser beam to be perfectly reflected, then the change in momentum of laser beam is known as $\dfrac{{2E}}{c}.$
Now, if we apply conservation of momentum, we can write it as,
$mv = \dfrac{{2E}}{c}$
$ \Rightarrow v = \dfrac{{2E}}{{mc}}$
As, we know that from conservation of energy, Kinetic energy = Potential energy
Therefore, we can write, $\dfrac{1}{2}m{v^2} = mg(1 - \cos \theta )$………… (i)
Also, from trigonometry, we know that,$1 - \cos \theta = 2{\sin ^2}\dfrac{\theta }{2}$
Now, since the angle is very small, we can write the above relation as, $1 - \cos \theta = 2{\left( {\dfrac{\theta }{2}} \right)^2}$=$\dfrac{{{\theta ^2}}}{2}$
Now, using the above relation in equation (i), we can write it as,
$\dfrac{1}{2}m{\left( {\dfrac{{2E}}{{mc}}} \right)^2} = mgl\dfrac{{{\theta ^2}}}{2}$
$ \Rightarrow {\theta ^2} = \dfrac{{2 \times 2{E^2}}}{{c{m^2}gl}}$
$\therefore \theta = \dfrac{{2E}}{{mc\sqrt {gl} }}$
As we can notice, the required value of the angle is not given in the options of the given question.
Hence, option (D), i.e. none of the above is the correct choice of the given question.
Note: We define a laser beam as a ray of light with high intensity that is generated through the emission of photons (also known as light particles). Photons do not have mass but they are capable of redirecting particles with which they collide and transfer their momentum to them.
Recently Updated Pages
JEE Main 2023 (April 8th Shift 2) Physics Question Paper with Answer Key

JEE Main 2023 (January 30th Shift 2) Maths Question Paper with Answer Key

JEE Main 2022 (July 25th Shift 2) Physics Question Paper with Answer Key

Classification of Elements and Periodicity in Properties Chapter For JEE Main Chemistry

JEE Main 2023 (January 25th Shift 1) Maths Question Paper with Answer Key

JEE Main 2023 (January 24th Shift 2) Chemistry Question Paper with Answer Key

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility & More

Class 11 JEE Main Physics Mock Test 2025

JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking

JEE Main 2023 January 24 Shift 2 Question Paper with Answer Keys & Solutions

Learn About Angle Of Deviation In Prism: JEE Main Physics 2025

JEE Main 2025: Conversion of Galvanometer Into Ammeter And Voltmeter in Physics

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements

NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Fluids

Units and Measurements Class 11 Notes: CBSE Physics Chapter 1

NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line

Important Questions for CBSE Class 11 Physics Chapter 1 - Units and Measurement
